"Algorithm notes" learning record 002

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First, today SLP

p14~27

Remarks:

1. I ++ ++ i and distinction of: i ++ is then used to i + 1; ++ i i + 1 is the first re-use i.
2. Infinite number of INF said:

   const int INF = (1<<30)-1;
   const int INF = 0x3fffffff;

3. Left to the right:

Specify A = 60 (0011 1100):

(1) bitwise left shift (<<) all bits leftward operand specified number of digits.

The figure below shows the results 11111111 << 1 (11111111 left one) is. Bit digital representation is moved blue, gray is discarded bits 0 vacancies filled with orange.

(A << 2) result 240, binary 1111 0000

(2) the right bit (<<) to move to the right all the bits of the operand specified number of digits.

The figure below shows the results (right one 11111111) 11111111 >> 1. Bit digital representation is moved blue, gray is discarded bits 0 vacancies filled with orange.

A >> 2 result 15, is binary 0000 1111

4. It is equivalent to the following two lines of code, but preferably parentheses. (Avoid mistakes unskilled)

   n/= m+1 ;
   n/=(m+1);

5. It does not require additional address operator & array name before. In scanf , except where char array (string) without & entire input, the other variable types need to add & .
6. The double scanf is% lf; printf is% f.
7. How% output and \:

   printf("%%");
   printf("\\"); 

8. printf的三种实用输出格式：%md %0md %.mf（m是常数）
9. “四舍六入五成双”

“四舍六入五成双”，也即“4舍6入5凑偶”，这里“四”是指≤4 时舍去，"六"是指≥6时进上。"五"指的是根据5后面的数字来定，当5后有数时，舍5入1；当5后无有效数字时，需要分两种情况来讲：5前为奇数，舍5入1；5前为偶数，舍5不进（0是偶数）。

一、具体计算规则：

（1）被修约的数字小于5时，该数字舍去；

（2）被修约的数字大于5时，则进位；

（3）被修约的数字等于5时，要看5前面的数字，若是奇数则进位，若是偶数则将5舍掉，即修约后末尾数字都成为偶数；若5的后面还有不为“0”的任何数，则此时无论5的前面是奇数还是偶数，均应进位。

二、来由：

这一方式的另一个常见名称为“银行家舍入”，是IEEE754标准的推荐舍入标准。这一方式跟通常的四舍五入相比，平均数方面更能保持原有数据的特性。

三、举例：

9.8249=9.82, 9.82671=9.83

9.8350=9.84, 9.8351 =9.84

9.8250=9.82, 9.82501=9.83

10. getchar可以识别换行符，输出的时候就会换一行。
    
11. 常用math函数：使用时程序前面要加头文件

  #include <math.h>

1.fabs(double x) //绝对值函数
2.floor(double x)和ceil(double x) //向下/向上取整
3.pow(double r,double p) //返回r^p
4.sqrt(double x) //返回算术平方根
5.log(double x) //返回自然对数，其他对数用换底公式
6.sin(double x)、cos(double x)和tan(double x) //参数要求是弧度制pi*45/180
7.asin(double x)、acos(double x)和atan(double x) 
8.round(double x) //四舍五入，返回类型是double型，强制转换取整

12. 角度制与弧度制

角度制与弧度制

1. 角度制（Degree Measure）
   
   把一个圆周平均分成360份，其中的每一份都是1o的角。这种以“度”作为单位来度量角度单位制叫做角度制。下图是我们常见的180o角度尺Protractor。

2. Radians (Radian Measure)
   
   length of long radius arc, the central angle of the arc is 1 (Radian), indicated by symbol rad.
   Is a positive angle in radians positive and negative angle of curvature is a negative number, the number zero angle in radians. Long radius arc of a circle the center of the angle α r is l, then the absolute value of [alpha] is the angle in radians | [alpha]
   | = L / R & lt.

3. The angle in radians and converted
   
   common control angle in radians FIG.
   

13. Exact value of pi π = acos (-1.0)
14. % F default retention six decimal.

Second, today's code analysis

2.02

#include <stdio.h>
int main(){
	char c1,c2,c3;
	c1=getchar();
	getchar();
	c2=getchar();
	c3=getchar();
	putchar(c1);
	putchar(c2);
	putchar(c3);
	return 0;
}

b where to go?
- declare three variables, output three variables. getchar is connected character ah, a friend, a received, not assignment out, not to suddenly disappear. c1 receiving a; getchar () receives b, thus saving not suddenly disappear; C3 received c; c4 receiving d.

Third, practice 2.2

2.03A

#include <stdio.h>
int main(){
	printf("This is my first c program!");
	return 0;
}

2.04B

#include <stdio.h>
int main(){
	printf("********************\nVery Good!\n********************");
	return 0;
}

2.05C

#include <stdio.h>
int main(){
	int a=123;
	int b=456;
	int sum=a+b;
	printf("sum=%d",sum);
	return 0;
}

2.06D

#include <stdio.h>
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	printf("%d",a+b);
	return 0;
}

2k07a

#include <stdio.h>
#include <math.h>
int main(){
	double a,b,c,r1,r2;
	scanf("%lf%lf%lf",&a,&b,&c);
	if(a=0||b*b-4*a*c<=0){
		printf("a=0或b^2-4ac不大于0，请重新输入参数！"); 
	}
	else{
		r1 = ((-b) + sqrt(b*b - 4*a*c))/(2*a);	
		r2 = ((-b) - sqrt(b*b - 4*a*c))/(2*a);
		printf("r1=%.2f\n",r1);	
		printf("r2=%.2f",r2);	
	}
	return 0;
}

#include <stdio.h>
#include <math.h>
int main(){
	double a,b,c,r1,r2;
	scanf("%lf%lf%lf",&a,&b,&c);
	if(b*b-4*a*c<=0){
		printf("a=0或b^2-4ac不大于0，请重新输入参数！"); 
	}
	else{
		r1 = ((-b) + sqrt(b*b - 4*a*c))/(2*a);	
		r2 = ((-b) - sqrt(b*b - 4*a*c))/(2*a);
		printf("r1=%.2f\n",r1);	
		printf("r2=%.2f",r2);	
	}
	return 0;
}

He has changed a bit, although they are still trying to find fault.

#include <stdio.h>
#include <math.h>
int main(){
	double a,b,c,r1,r2;
	scanf("%lf%lf%lf",&a,&b,&c);
	if(a!=0||b*b-4*a*c>0){
		r1 = ((-b) + sqrt(b*b - 4*a*c))/(2*a);	
		r2 = ((-b) - sqrt(b*b - 4*a*c))/(2*a);
		printf("r1=%.2f\n",r1);	
		printf("r2=%.2f",r2);	
	}
	else	
	{
		printf("a=0或b^2-4ac不大于0，请重新输入参数！"); 
	}
	return 0;
}

2.08F

#include <stdio.h>
#include <math.h>
int main(){
	char str[10];
	scanf("%s",str);
	printf("%s\n",str);	
	return 0;
}